Question: Assume that $S$ is an inwardly oriented, piecewise-smooth surface with a piecewise-smooth, simple, closed boundary curve $C$ oriented positively with respect to the orientation of $S$. The boundary curve $C$ makes a right angle at the origin. Let $F$ be a continuously differentiable vector field in $\mathbb{R}^3$. Does Stokes' theorem necessarily apply to the surface $S$, boundary curve $C$, and vector field $F$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Explanation: Assume we have a continuously differentiable three-dimensional vector field $F(x, y, z)$, an oriented piecewise-smooth surface $S$, and a piecewise-smooth, simple, closed boundary curve $C$ oriented positively with respect to $S$. Then Stokes' theorem states that we have the equality below: $ \oint_C F \cdot dr = \iint_S \text{curl}(F) \cdot dS$ If $C$ is negatively oriented, the line integral is equal to the negative of the double integral. [What does any of that mean?] We satisfy every condition required for Stokes' theorem, but what about the right angle that $C$ makes at the origin? It actually doesn't matter, because we only require that $C$ is piecewise-smooth. Imagine chopping $C$ in half at the origin. We would be left with two smooth curves, and their combination is therefore piecewise-smooth. Therefore, the answer is yes, Stokes' theorem does apply in this case.